# How do you find the domain and range of f(x) = sqrt(4 - x²)?

Jul 2, 2015

Domain: $- 2 \le x \le 2$;
Range: $0 \le y \le 2$

#### Explanation:

You want to avoid to have a negative number as argument of your square root, so you must set:
$4 - {x}^{2} \ge 0$ that can be rearranged to get (changing signs):
${x}^{2} \le 4$
$x \le \pm \sqrt{4}$
$x \le \pm 2$
So, basically, your function can accept only $x$ values inside the interval from $- 2$ to $+ 2$ giving a domain: $- 2 \le x \le 2$.
The maximum $y$ value of your function is reached when $x = 0$ and corresponds to $y = 2$. So the range will be from $0$ to $+ 2$ or: $0 \le y \le 2$

Graphically:
graph{sqrt(4-x^2) [-10, 10, -5, 5]}